Question

Can someone show me how to turn this parabola into standard form?

y^2 + 4y - 8x + 4 = 0

Answer 1

I'm not quite sure what the standard form of a parabola is

Answer 2

standard form looks like
\[y = ax^2+bx+c\]for a parabola that opens up or down
and
\[x = ay^2+by+c\]for a parabola that opens right or left

Answer 3

So it would be 8x = y^2 + 4y + 4?

Answer 4

do you know what a "perfect square trinomial" is ?

Answer 5

You asked me that in my last question and I think I do know what it is but you gave me a youtube video and I can't wait videos on this computer.

Answer 6

*watch

Answer 7

well, the video shows.... how to turn it into it =) thus

Answer 8

My computer doesn't play videos though /:

Answer 9

?

Answer 10

http://images.flatworldknowledge.com/redden/redden-eq05_089.jpg

Answer 11

\[8x=y^2+4y+4\]\[x=\frac{1}{8}(y^2+4y+4)\]Is your standard form.

Now can you write the stuff in () as a square?

Answer 12

The stuff in parenthesis as a square would be (y + 2)^2, right?

Answer 13

so x = 1/8(y+2)^2 ?

Answer 14

\(\bf y^2 + 4y - 8x + 4 = 0\implies 8x=(y^2 + 4y + 2^2)\\ \quad \\
\implies 8x=(y+2)^2\implies 8(x-0)=(y+2)^2\)

Answer 15

Where did the 0 come from?

Answer 16

8x = 8x+0 or 8x-0 = 8x

Answer 17

oh okay

Answer 18

It can't be left at x = 1/8 (y + 2)^2 ?

Answer 19

sure you can, then it won't be standard form per se

Answer 20

I just went through my notes and my equation for horizontal parabolas is x = 1/4p y^2

Answer 21

So I think that's the most my teacher expects from me for now lol

Answer 22

Thank you!

Answer 23

hmmm ahemm well.. you see... for a parabola either form will do
\(\bf (y-k)^2=4p(x-h)\qquad x=ay^2+by+c\)

are both valid standard forms of a parabola

if you want the latter, just solve for "x" like whpalmer4 showed above

if you want the former, then use that :)

Answer 24

rearranging it as a square as you did is also the first step to putting it in "vertex" form:

\[y = (x-h)^2+k\]where you can just read off the coordinates of the vertex as \((h,k)\)